The TV show Numb3rs and probability

Ok you gotta admit this show is cool but I don't always know if this guy is talking math or if someone is just making this math up but, well he gave a wierd probability example that doesn't make sense..

You are at a gameshow and there are three curtains before you. Behind one curtain is a goat, behind the other curtain is a car, and behind the other is another goat.

The object is to guess the curtain which has the car behind it, and you win it.

Now obviously the chances on the first guess, are 1/3. (One choice three possibilities.)

Now a contestent pics the center curtain. A side curtain is opened, revealing a goat. So one curtain is eliminated and you know that the car is either behind the curtain you picked, or the other curtain. NOW will switching your choice improve your chances of winning the car? It would appear, that either way, your chances are the same. 50-50. But according to Mr Charlie Eps, you'd be wrong. Unfortunatly he explained why so quickly I couldn't follow it. But he said that switching your choice at that point, actually doubles your chances of winning the car. Something about revealing a goat in the other curtain increases the odds of a goat being in the location you picked. Why, I have no idea, I couldn't follow it that quickly.

Has anyone ever heard of this problem? Can you justify why switching your choice after the goat is revealed in the other curtain, doubles your chances?

Re: The TV show Numb3rs and probability

I think that you should switch because they didn't open the goat curtain at random, they may have specificallyhave chosen the goat, not the car behind the other curtain. I think I heard this long ago, some questionanalogous to this one.

Re: The TV show Numb3rs and probability

Yes, it is a known puzzle, and numb3rs seems to have got it right (though I haven't seen that episode).

Perhaps this fact is important: the host does not reveal one door at random - he won't reveal the contestants door of course, and he won't reveal a door that has a car.

If the contestant sticks with his original choice he will continue to have a 1-in-3 chance, so now there are only two doors left, the other must have a 2-in-3 chance. Wierd but true when you think about it.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Re: The TV show Numb3rs and probability

Sorry all you people who think this works. IT DOESN'T!!

Proof

The past will never ever ever determine a greater probability for one posibility. The fact that you made a decision before does not change the probability that you have 2 options, "swap" or "stay" and two posibilities "car" or "goat". The chance is 50/50 (FULLSTOP).

Imagine that you dont choose one to begin with!! Lets play this out. options A B C. They tell you to pick one (BUT YOU DONT). Knowing that they will not pick the "car" they show you that C is a goat. Then they tell you to "stay" or "swap". You dont have a orginal choice. SO WHATS THE PROBABILITY!!! You have to choose A or B. 1 is right, 1is wrong! The probability is 50:50. The same as if they never had a C in the first place.

If You still dont belive me then believe me when I say that Im a computer Scientist and I simulated the whole process in a computer and run the simulation ONE HUNDRED MILLION TIMES!!!!! The simulation showed that it swapping worked 50% of the times.

Re: The TV show Numb3rs and probability

100 000 000 times yes!!!

and I did that a few times, so really about 400 000 000 times

and yes "The past will never ever ever determine a greater probability for one posibility"eg.if you throw a dice 9 times, and a 6 comes up every single time (highly unlikely, but possible). What would be the probability of a 6 coming up on your tenth throw? ANSWER 1/6. Ive also run this simulation . Even though the chance that of throwing a 6, 10 times in a row is 1 in 60466176 if it has already been thrown 9 times, the next time is 1/6.

Its the same principle

So the probability started at 1-in-3 but then the host revealed one, the probability changes to 50:50.

Re: The TV show Numb3rs and probability

I've seen an example where there are 100 doors with 1 car and 99 goats.

Now, you choose a door, then the host reveals 98 doors which do not contain the car.This leaves 2 doors: the door you chose and 1 other.

The probablity of initially choosing the door with the car behind it is 1 in 100.

On its own the above statement is true. However, we know for a FACT thatthe host will always narrow the doors down to 2, while always avoiding the door with the car behind.

With this added information the statement now becomes false. Why?

Because we will always end up with 2 doors, 1 with the car behind it, and 1 with a goat behind it.Your initial choice is irrelevant.

Thus, the probability of initially choosing the door with the car behind it,with the knowledge that the host will reveal 98 incorrect doors, is 1 in 2!

The irony of the question in the original post is any fool will say 50/50 for their first answer... and be correct!

It is to be noted that this only works because of the rules of the game.Were it random what doors were revealed, it would be an entirely different situation.In this case, the car could be revealed at any time and every time a car is not revealed increases the chance that the door you initially chose contains the car.

Anyway, the trick is not to dwell on it too much, or you may find yourself looking for answers which do not exist!

Re: The TV show Numb3rs and probability

Sorry all you people who think this works. IT DOESN'T!!

Proof

The past will never ever ever determine a greater probability for one posibility. The fact that you made a decision before does not change the probability that you have 2 options, "swap" or "stay" and two posibilities "car" or "goat". The chance is 50/50 (FULLSTOP).

Imagine that you dont choose one to begin with!! Lets play this out. options A B C. They tell you to pick one (BUT YOU DONT). Knowing that they will not pick the "car" they show you that C is a goat. Then they tell you to "stay" or "swap". You dont have a orginal choice. SO WHATS THE PROBABILITY!!! You have to choose A or B. 1 is right, 1is wrong! The probability is 50:50. The same as if they never had a C in the first place.

If You still dont belive me then believe me when I say that Im a computer Scientist and I simulated the whole process in a computer and run the simulation ONE HUNDRED MILLION TIMES!!!!! The simulation showed that it swapping worked 50% of the times.

CONCLUSION:NUMB3RS WAS WRONG!!

Something doesn't sound right there. You eliminated the selection. In your example, you said we will not make an official selection and reveal goat C every time. Am I correct? Pardon me if I'm not but If I am, what if you were to pick goat C when actually playing the game? The rules of the game state that when you pick one, one of the goats will be revealed, not the car, and not the one you picked. So if you pick a a door, and he flips over a goat. Now there are three possibilities at this point. You picked goat A so he has to reveal goat B. Or you picked goat B so he has to reveal got A. Or you picked the Car and he randomly flipped over one of the goats. In the first two situations, switching your choice would win you the car since its always more likely he is revealing a remaining goat. In the third situation, if you picked the car on the first try, switching would loose you the car. So in 2 out of 3 situations, switching will win you the car.

I think its because no matter which card you pick, there will always be one remaining goat that is not the door you picked (since there are two goats). So the revelation does not improve your chances at all. So if you pick door B and he reveals door C. It is most likely he revealed door C rather then door A because door A contains the car. Its not that the previous chances count, its that your chances have not improved because there will always be a remaining goat to reveal, and that will always be the one revealed. So your chances have not improved, which means they haven't changed which means they are now what they were then.

I read about computer simulations that proved this concept. Did you program it to make the selection, reveal a remaining goat that is NOT the one you picked, switch the choice, and check whats inside? Thats the only proper way to do it.

Re: The TV show Numb3rs and probability

I made a post that pretty much echoed Mikau's, but when I posted it, I saw his and deleted it again. The chance is most definately 1/3, though.

And to the person who said that the past doesn't change future events, that's only when they are independant.Some events, like the game show, depend on what has happened already to decide what will happen next. For example, if you pick a card from a deck of cards, the chance of picking it again is 0, because it wouldn't be in the deck anymore.

Re: The TV show Numb3rs and probability

If You still dont belive me then believe me when I say that Im a computer Scientist and I simulated the whole process in a computer and run the simulation ONE HUNDRED MILLION TIMES!!!!! The simulation showed that it swapping worked 50% of the times.

Re: The TV show Numb3rs and probability

Something doesn't sound right there. You eliminated the selection. In your example, you said we will not make an official selection and reveal goat C every time. Am I correct?

C is just simply one that is not the car.

mikau wrote:

Pardon me if I'm not but If I am, what if you were to pick goat C when actually playing the game? The rules of the game state that when you pick one, one of the goats will be revealed, not the car, and not the one you picked.

Ok, lets just say that B "is" the car. Just to clarifyIf I picked A then they would reviel CIf I picked C then they would reveil A

So you see it didn't matter whether I picked A or C in the first place. Im still in the same predicament. And obviously, if Id have picked B in the first place, then then I swap, I'd be wrong.

mikau wrote:

I read about computer simulations that proved this concept. Did you program it to make the selection, reveal a remaining goat that is NOT the one you picked, switch the choice, and check whats inside? Thats the only proper way to do it.

Re: The TV show Numb3rs and probability

Why did you not get a total of 100 million? Instead your total is roughly 2/3 of 100 million. 1/3 of 100 million worked, 1/3 of 100 million did not work, what about the other third? If it goes to the switch then numb3rs WAS right.

What did you mean by "Makes sure the one reveiled was not the answer, if it was, the whole thing is discarded." Discarded? I'm not sure what you mean but just to make sure, after (for instance) door A is picked. The program should check to see whats behind door B, if its a goat, then reveal it and switch the choice to door C. If you picked door A and door B has a car behind it, then B cannot be revealed so you reveal door C and switch your choice WHICH WILL WIN YOU THE CAR! If you simply quite the game if theres a car behind the door you want to reveal, then about 1/3 of the answers should be missing, and that third should go to the switch method.

Re: The TV show Numb3rs and probability

I know some basic C++ but I don't know much about arrays or static whatevers. lol. So I only half understand how that program works. But 1/3 of the answers are missing so somethings not right. If you flip a coin 100,000 million times, it should come up heads about 50 million times, and tails about 50 million times. If you end up with 33 million heads and 33 million tails then someone forgot to mark something down. (about 33 million times! :-O)

It cannot be denied, if the rules of the game are followed, if you do not pic a goat on the first time, you will win if you switch. The odds of picking a goat are 2/3. So there you have it.

Re: The TV show Numb3rs and probability

I know why 1/3 of the answers are missing, and its cause I was lazy. Rather than finding a door which wasn't the guess and which wasn't the answer, I simply picked a random number and if it wasn't the guess and wasn't the answer then I started from the top at the next itteration

shouldn't make a difference, but I try and go back and be more through if you like. I be back with modified code

Re: The TV show Numb3rs and probability

it will make a differance al right. that will happen about 1/3 of the time, and when it does, the point will go to the switcher. So that 1/3 of the points that are missing, all go to the switcher. That brings the odds up to 2/3.

The thing that I couldn't get my head around was, if you hadn't choosen one in first place. I couldn't work out what was the difference. Now I realise the difference is that you can choose one card that you dont want to be eliminated as being a goat. And by restricting the host to what he can remove thus it increses you chances on the swap.